Question

A uniform rod of mass $m$ and length $l$ is in equilibrium under the action of constraint forces, gravity and tension in the string. Find the
(a) frictional force acting on the rod
(b) tension in the string
(c) normal reaction on the rod
Now the string is cut. Find the
(d) angular acceleration of the rod just after the string is cut
(e) normal reaction on the rod just after the string is cut

Answer 1

Resolving the forces parallel and perpendicular to rod.

Along the length of rod.

$f=mg\sin {60}^o=\frac{\sqrt{3mg}}{2}$

Hence friction force action on rod at support $P$ will be $mg/2$

(b) for finding tension we can take torque about support $P$

$T\left(\frac{3}{4}l\right)=mg\left(\frac{l}{4}\right)orT=\frac{mg}{3}$

(c) Just before cutting the string the rod is at equilibrium. Considering the forces perpendicular to rod length

$N+T=mg\cos {60}^o\Rightarrow N=\frac{mg}{2}-\frac{mg}{3}=\frac{mg}{6}$

(d) Just after cutting the string the equilibrium of the rod will disturbed. The rod will have angular acceleration.

As the rod does not slip point $P$ will be at rest at the time just after cutting the string

We can apply torque equation about ${\tau }_P=I_P\alpha$

$\left(mg\cos {60}^o\right)\frac{l}{4}=(\frac{m{l}^2}{12}+m{\left(\frac{l}{4}\right)}^2)\alpha \Rightarrow \alpha =\frac{6g}{7l}$

(e) If we apply torque equation about centre of mass

$N.\frac{l}{4}=\left(\frac{m{l}^2}{12}\right)\alpha$

$N.\frac{l}{4}=\left(\frac{m{l}^2}{12}\right)\left(\frac{6g}{7l}\right)\Rightarrow N=\frac{2}{7}mg$